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      集合元素个数与线性空间的维数
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          <p>证明一个集合元素个数不超过线性空间的维数, 可以去证明这个集合的元素是线性无关的. 而证明线性空间的维数不超过集合元素个数n, 可以考虑证明线性空间中任意n+1个元素都线性相关. 这在域论中是常见的手法.</p>
<p>下面给出几个例子.</p>
<a id="more"></a>
<blockquote>
<p><strong>问题 1</strong> 设<span class="math inline">\(V\)</span>为 n 维实线性空间, <span class="math inline">\(M\subseteq\mathrm{End}V\)</span>, 满足</p>
<ol type="1">
<li><p><span class="math inline">\(\mathrm{id}\in M,0\notin M\)</span>;</p></li>
<li><p>若<span class="math inline">\(\mathscr{A},\mathscr{B}\in M\)</span>, 则<span class="math inline">\(\mathscr{AB}\in M\)</span>或<span class="math inline">\(\mathscr{BA}\in M\)</span>;</p></li>
<li><p>若<span class="math inline">\(\mathscr{A},\mathscr{B}\in M\)</span>, 则<span class="math inline">\(\mathscr{AB}=\mathscr{BA}\)</span>或<span class="math inline">\(\mathscr{AB}=-\mathscr{BA}\)</span>;</p></li>
<li><p>若<span class="math inline">\(\mathscr{A}\in M\)</span>且<span class="math inline">\(\mathscr{A}\neq\pm\mathrm{id}\)</span>, 则存在<span class="math inline">\(\mathscr{B}\in M\)</span>, s.t. <span class="math inline">\(\mathscr{AB=-BA}\)</span></p></li>
</ol>
<p>证明: <span class="math inline">\(M\)</span>中的元素个数不超过<span class="math inline">\(2n^{2}\)</span>.</p>
</blockquote>
<p>证明: (提示: 证明一个集合元素个数不超过线性空间的维数, 可以去证明这个集合的元素是线性无关的. 此处<span class="math inline">\(n^{2}\)</span>暗示维数, <span class="math inline">\(\mathscr{A},-\mathscr{A}\)</span>可以同时出现, 所以有个 2 倍.)</p>
<p>首先证明<span class="math inline">\(\forall\mathscr{A}\in M\)</span>, 都有<span class="math inline">\(\mathscr{A}^{2}=\pm\mathrm{id}\)</span>.</p>
<p>任取<span class="math inline">\(\mathscr{A}\in M\)</span> , 由(2)得<span class="math inline">\(\mathscr{A}^{2}\in M\)</span>. 如果<span class="math inline">\(\mathscr{A}^{2}\neq\pm\mathrm{id}\)</span> , 则由(4), <span class="math inline">\(\exists\mathscr{B}\in M\)</span>, s.t. <span class="math inline">\(\mathscr{A}^{2}\mathscr{B}=-\mathscr{BA}^{2}\)</span>.</p>
<p>另一方面, 若<span class="math inline">\(\mathscr{AB}=\mathscr{BA}\)</span>, 则<span class="math inline">\(\mathscr{A}^{2}\mathscr{B}=\mathscr{A}(\mathscr{AB})=\mathscr{A}(\mathscr{BA})=(\mathscr{AB})\mathscr{A}=\mathscr{B}\mathscr{A}^{2}\)</span>; 若<span class="math inline">\(\mathscr{AB}=-\mathscr{BA}\)</span>, 则<span class="math inline">\(\mathscr{A}^{2}\mathscr{B}=\mathscr{A}(\mathscr{AB})=\mathscr{A}(-\mathscr{BA})=-(\mathscr{AB})\mathscr{A}=\mathscr{B}\mathscr{A}^{2}\)</span>. 即总有<span class="math inline">\(\mathscr{A}^{2}\mathscr{B}=\mathscr{B}\mathscr{A}^{2}\)</span>. 所以<span class="math inline">\(M\ni\mathscr{A}^{2}\mathscr{B}=0\)</span>, 这与(1)矛盾.</p>
<p>然后证明<span class="math inline">\(M\)</span>中不同且不互为相反数的元素线性无关.</p>
<p>(反证法)假设<span class="math inline">\(M\)</span>中不同且不互为相反数的元素线性相关, 那么一定可以找到一个最小的<span class="math inline">\(n\)</span>, 使得<span class="math inline">\(M\)</span>中不同且不互为相反数的<span class="math inline">\(n\)</span>个元素线性相关.</p>
<p>设<span class="math inline">\(\mathscr{A}_{1}, \mathscr{A}_{2}, \cdots, \mathscr{A}_{n}\in M\)</span>互不相同且<span class="math inline">\(\forall i\neq j\)</span>, <span class="math inline">\(\mathscr{A}_{i}+\mathscr{A}_{j}\neq0\)</span>, <span class="math inline">\(\mathscr{A}_{1}, \mathscr{A}_{2}, \cdots, \mathscr{A}_{n}\)</span>线性相关.</p>
<p><span class="math inline">\(\exists k_{i}\neq0\)</span> , 使得 <span class="math inline">\(\sum_{i=1}^{n} k_{i} \mathscr{A}_{i}=0\)</span>. 不妨设<span class="math inline">\(\mathscr{A}_{1}=\pm\mathrm{id}\)</span> (否则考虑<span class="math inline">\(\sum_{i=1}^{n} k_{i} \mathscr{A}_{1} \mathscr{A}_{i}=0\)</span>) , 则<span class="math inline">\(\mathscr{A}_{2},\cdots,\mathscr{A}_{n}\neq\pm\mathrm{id}\)</span>.</p>
<p><span class="math inline">\(\mathscr{A}_{n}\neq\pm\mathrm{id}\)</span>和(4)可知<span class="math inline">\(\mathscr{C}\in M\)</span>, s.t. <span class="math inline">\(\mathscr{A}_{n}\mathscr{C}=-\mathscr{C}\mathscr{A}_{n}\)</span>.</p>
<p>不妨假设对<span class="math inline">\(1\leq i\leq t\)</span>, <span class="math inline">\(\mathscr{A}_{i}\mathscr{C}=\mathscr{C}\mathscr{A}_{i}\)</span>; 对<span class="math inline">\(t+1\leq i\leq n-1\)</span>, <span class="math inline">\(\mathscr{A}_{i}\mathscr{C}=-\mathscr{C}\mathscr{A}_{i}\)</span>. 对式子</p>
<p><span class="math display">\[
\sum_{i=1}^{n-1}k_{i}\mathscr{A}_{i}=-k_{n}\mathscr{A}_{n}
\]</span></p>
<p>分别用<span class="math inline">\(\mathscr{C}\)</span>左作用和右作用, 得</p>
<p><span class="math display">\[
\sum_{i=1}^{t}k_{i}\mathscr{C}\mathscr{A}_{i}+\sum_{i=t}^{n-1}k_{i}\mathscr{C}\mathscr{A}_{i}=-k_{n}\mathscr{C}\mathscr{A}_{n}
\]</span></p>
<p><span class="math display">\[
\sum_{i=1}^{t}k_{i}\mathscr{A}_{i}\mathscr{C}+\sum_{i=t}^{n-1}k_{i}\mathscr{A}_{i}\mathscr{C}=-k_{n}\mathscr{A}_{n}\mathscr{C}
\]</span></p>
<p>由于对<span class="math inline">\(t+1\leq i\leq n\)</span>, <span class="math inline">\(\mathscr{A}_{i}\mathscr{C}=-\mathscr{C}\mathscr{A}_{i}\)</span>. 故将两式相加, 得到</p>
<p><span class="math display">\[
\sum_{i=1}^{t}k_{i}\mathscr{C}\mathscr{A}_{i}=0
\]</span></p>
<p>而<span class="math inline">\(t\leq n-1\)</span>, 与<span class="math inline">\(n\)</span>的最小性矛盾.</p>
<p>因此, <span class="math inline">\(M\)</span>中不同且不互为相反数的元素线性无关. 而<span class="math inline">\(\dim M\leq n^{2}\)</span>. 故<span class="math inline">\(\left|M\right|\leq2n^{2}\)</span>. <span class="math inline">\(\square\)</span></p>
<blockquote>
<p><strong>问题 2</strong> [Dedekind-Artin] 设<span class="math inline">\(G\)</span>是一个幺半群, <span class="math inline">\(\mathbb{K}\)</span>是一个域 (则<span class="math inline">\(K^{*} := \mathbb{K}-\{0\}\)</span>是一个群). <span class="math inline">\(\sigma_{1},\sigma_{2},\ldots,\sigma_{n}\)</span>是两两不同的非零同态<span class="math inline">\(G\rightarrow K^{*}\)</span>, 则它们在<span class="math inline">\(\mathbb{K}\)</span>上线性无关.</p>
</blockquote>
<p>证明: 假设存在这样的一组非零同态, 使得它们在<span class="math inline">\(\mathbb{K}\)</span>上线性无关, 则一定能找到其中元素个数最少的一组. 设<span class="math inline">\(n\)</span>是满足</p>
<p><span class="math display">\[
a_{1}\sigma_{1}+\cdots+a_{n}\sigma_{n}=0,\ a_{i}\in\mathbb{K}\text{不全为}0
\]</span></p>
<p>的最小的正整数. 则<span class="math inline">\(n\geq2\)</span>, <span class="math inline">\(a_{i}\)</span>均不为 0.</p>
<p>因为<span class="math inline">\(\sigma_{1},\sigma_{2}\)</span>不同, 故<span class="math inline">\(\exists z\in G\)</span>使得<span class="math inline">\(\sigma_{1}(z)\neq\sigma_{2}(z)\)</span>. 对于任意<span class="math inline">\(x\in G\)</span>, 都有 <span class="math display">\[
a_{1}\sigma_{1}(xz)+\cdots+a_{n}\sigma_{n}(xz)=0
\]</span></p>
<p>由于<span class="math inline">\(\sigma_{i}\)</span>是同态, 则有</p>
<p><span class="math display">\[
a_{1}\sigma_{1}(z)\sigma_{1}+\cdots+a_{n}\sigma_{n}(z)\sigma_{n}=0
\]</span></p>
<p>两边同除<span class="math inline">\(\sigma_{1}\)</span>并与第一个式子相减, 得</p>
<p><span class="math display">\[
\left(a_{2}\dfrac{\sigma_{2}(z)}{\sigma_{1}(z)}-a_{2}\right)\sigma_{2}+\cdots+\left(a_{n}\dfrac{\sigma_{n}(z)}{\sigma_{1}(z)}-a_{n}\right)\sigma_{n}=0
\]</span></p>
<p>其第一个系数就不为 0, 且比第一个式子少一个元素, 这与<span class="math inline">\(n\)</span>的最小性矛盾.</p>
<p>因此, 任意一组两两不同的非零同态<span class="math inline">\(G\rightarrow K^{*}\)</span>在<span class="math inline">\(\mathbb{K}\)</span>上线性无关. <span class="math inline">\(\square\)</span></p>
<blockquote>
<p><strong>问题 3</strong> [Artin] 设<span class="math inline">\(\mathbb{E}\)</span>为数域, <span class="math inline">\(G\)</span>为<span class="math inline">\(\mathrm{Aut}\mathbb{E}\)</span>的有限子群, 则<span class="math inline">\(\left|G\right|\geq\left[\mathbb{E}:\mathbb{F}\right]\)</span>.</p>
</blockquote>
<p>证明: 设<span class="math inline">\(G=\{\sigma_{1}=\mathrm{id},\sigma_{2},\ldots,\sigma_{n}\}\)</span>. 要证<span class="math inline">\(\left[\mathbb{E}:\mathbb{F}\right]\leq\left|G\right|=n\)</span>, 只要证<span class="math inline">\(\mathbb{E}\)</span>上任意<span class="math inline">\(n+1\)</span>个元素在<span class="math inline">\(\mathbb{F}\)</span>上线性相关.</p>
<p><span class="math inline">\(\forall\alpha_{1},\alpha_{2},\ldots,\alpha_{n},\alpha_{n+1}\in\mathbb{E}\)</span>, 则<span class="math inline">\(\mathbb{E}\)</span>上的线性方程组</p>
<p><span class="math display">\[
\begin{pmatrix}\sigma_{1}(\alpha_{1}) &amp; \sigma_{1}(\alpha_{2}) &amp; \cdots &amp; \sigma_{1}(\alpha_{n+1})\\
\sigma_{2}(\alpha_{1}) &amp; \sigma_{2}(\alpha_{2}) &amp; \cdots &amp; \sigma_{2}(\alpha_{n+1})\\
\vdots &amp; \vdots &amp;  &amp; \vdots\\
\sigma_{n}(\alpha_{1}) &amp; \sigma_{n}(\alpha_{2}) &amp; \cdots &amp; \sigma_{1}(\alpha_{n+1})
\end{pmatrix}_{n\times(n+1)}\begin{pmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n+1}
\end{pmatrix}=0
\]</span></p>
<p>必有非 0 解 (未知数的个数大于方程个数).</p>
<p>考虑其中包含非 0 元素最少的非零解<span class="math inline">\(\begin{pmatrix}b_{1} &amp; b_{2} &amp; \cdots &amp; b_{m} &amp; 0 &amp; \cdots &amp; 0\end{pmatrix}\)</span>, 其中<span class="math inline">\(b_{i}\neq0,i=1,2,\ldots,m\)</span>. 不妨设<span class="math inline">\(b_{1}=1\)</span>, (否则考虑<span class="math inline">\(\begin{pmatrix}1 &amp; \dfrac{b_{2}}{b_{1}} &amp; \cdots &amp; \dfrac{b_{m}}{b_{1}} &amp; 0 &amp; \cdots &amp; 0\end{pmatrix}\)</span>).</p>
<p>方程组两边用<span class="math inline">\(\sigma_{i}\)</span>作用, 由<span class="math inline">\(\sigma_{i}\)</span>是同态, 可以得到<span class="math inline">\(\begin{pmatrix}\sigma_{i}(b_{1}) &amp; \sigma_{i}(b_{2}) &amp; \cdots &amp; \sigma_{i}(b_{m}) &amp; 0 &amp; \cdots &amp; 0\end{pmatrix}\)</span>是一组非零解, 其中<span class="math inline">\(\sigma_{i}(b_{1})=\sigma_{i}(1)=1\)</span>. 将两组解相减, 得</p>
<p><span class="math display">\[
\begin{pmatrix}0 &amp; \sigma_{i}(b_{2})-b_{2} &amp; \cdots &amp; \sigma_{i}(b_{m})-b_{m} &amp; 0 &amp; \cdots &amp; 0\end{pmatrix}
\]</span></p>
<p>也是一组非零解. 但这组解比<span class="math inline">\(\begin{pmatrix}b_{1} &amp; b_{2} &amp; \cdots &amp; b_{m} &amp; 0 &amp; \cdots &amp; 0\end{pmatrix}\)</span>含有更少的非零元, 因此它只能是零解, 即</p>
<p><span class="math display">\[
\sigma_{i}(b_{j})=b_{j},\ \forall1\leq i\leq n,\ 1\leq j\leq m
\]</span></p>
<p>由<span class="math inline">\(\mathbb{F}\)</span>的定义知<span class="math inline">\(b_{j}\in\mathbb{F},\ \forall1\leq j\leq m\)</span>. 所以</p>
<p><span class="math display">\[
0=\sum_{j=1}^{n+1}\sigma_{1}(\alpha_{j})b_{j}=\sum_{j=1}^{n+1}\mathrm{id}(\alpha_{j})b_{j}=\sum_{j=1}^{n+1}b_{j}\alpha_{j}
\]</span></p>
<p>这表明<span class="math inline">\(\alpha_{1},\ldots,\alpha_{n+1}\)</span>在<span class="math inline">\(\mathbb{F}\)</span>上线性相关. <span class="math inline">\(\square\)</span></p>

          
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